# Seeking clarification of Lebesgue definition given for $\int _{0}^{1}x^{-a}dx$

I came across the example "Show that $\int _{0}^{1}x^{-a}dx$ exists as a Lebesgue integral, and is equal to $1/(1-a)$, if $0 < a < 1$; but is infinite if $a\geq 1$. The Lebesgue definition of the integral is $$\lim _{n\rightarrow \infty }\left\{ \int _{0}^{n^{-\frac {1} {a}}}ndx+\int _{n^{-\frac {1} {a}}}^{1}x^{-a}dx\right\}$$ and the results are the same as in the elementary theory.

To put my question bluntly i just do not understand why and how the author determined to split the original integral. I am aware if we take the limit the first integral's upper bound would become 0 and the second integral's lower limit would be 0 and the second integral would look the same as the one we were originally presented with. I suppose i do not quite understand the motivation behind the step. Any light shed on this matter would be much appreciated.

Edit: As per request the definition provided in the book is.

The Lebesgue integral of $f(x)$ over $(a, b)$ is the common limit of the sums $s$ and $S$ when the number of division-points $y_v$ is increased indefinitely, so that the greatest value of $y_{v+1} - y_v$ tends to zero. where $$s=\sum _{v=0}^{n}y_{v}\mu\left( e_{\nu }\right)$$ and

$$S=\sum _{v=0}^{n}y_{v+1}\mu\left( e_{\nu }\right)$$

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The Lebesgue definition of the integral is NOT what you write. You might want to cite your source. –  Did Jul 14 '12 at 9:56
@did please see the edit. Thanks in advance. –  Hardy Jul 14 '12 at 10:08
This is not the definition of Lebesgue integrability. (And you did not give the title of the book.) –  Did Jul 14 '12 at 10:12
That is most certainly not the definition of the Lebesgue integral. It looks like the definition of the Riemann-Stieltjes integral. –  Potato Jul 14 '12 at 10:20
@did This is the definition from the book The Theory of Functions by E. Titchmarsh. Could please possibly share your definition ? so we can compare how they are different. –  Hardy Jul 14 '12 at 10:47

In modern parlance, Titchmarsh (The Theory of functions, Second edition, Section 10.7) defines the integral of an unbounded nonnegative measurable function $f$ as the limit of the integrals of $f_n=\min\{f,n\}$, where the integral of a bounded measurable function was defined earlier in the book along the lines of your Edit (which only applies to bounded functions, and not to the general case as one might be led to believe by your Edit, see Section 10.4 in the book).
Other, perhaps more convincing, treatments of Lebesgue integral exist, which yield the same result for the integral of $f$ than this one. Nevertheless, if one wants to stick to this definition, the job is to compute, in the case at hand, that is, for $f(x)=x^{-a}$ on $(0,1)$, $$\int_0^1f_n(x)\,\mathrm dx=n\cdot \left[x\right]_{x=0}^{x=n^{-1/a}}+\frac1{1-a}\cdot\left[x^{1-a}\right]_{x=n^{-1/a}}^{x=1}=\frac{1-a\cdot n^{1-1/a}}{1-a},$$ with the obvious modification when $a=1$, and to study the limit, if any, of the RHS when $n\to\infty$. To wit, each LHS is the integral of a bounded continuous function $f_n$ and, for these, one knows the result is the value of the Riemann integral.
Note that Lebesgue states the fact that the integral of $f\geqslant0$ is the limit of the integrals of $\min\{f,n\}$ but, if I remember correctly, to him this is a property, not a definition, of the integral of an unbounded nonnegative function. –  Did Jul 14 '12 at 12:13